Question

Sketch the solid S. Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ $$ S=\left\\{(x, y, z): 0 \leq x \leq \frac{1}{2} y, 0 \leq y \leq 4,0 \leq z \leq 2\right\\} $$

Short answer

\( \int_{y=0}^{4} \int_{x=0}^{\frac{1}{2}y} \int_{z=0}^{2} f(x, y, z) \, dz \, dx \, dy \)

Step-by-step solution

Analyze Boundaries

We start by analyzing the given boundaries. The region \( S \) is defined such that \( 0 \leq x \leq \frac{1}{2}y \), \( 0 \leq y \leq 4 \), and \( 0 \leq z \leq 2 \). This indicates a relationship between \( x \) and \( y \), suggesting the outer shape could be triangular in the \( xy \)-plane.

Visualize the Region in the XY-Plane

In the \( xy \)-plane, plot the boundary: \( x = \frac{1}{2}y \), which is a line passing through the origin with slope 1/2, extending until \( y = 4 \). The region is the set of points below this line from \( y = 0 \) to \( y = 4 \).

Extend Region to 3D

Consider the volume bounded between \( z = 0 \) and \( z = 2 \). This creates a prism-like solid that has a triangular cross-section in the \( xy \)-plane and is extended vertically from \( z = 0 \) to \( z = 2 \).

Set Up the Iterated Integral

To define the iterated integral for \( \iiint_{S} f(x, y, z) \, dV \), a logical approach is to choose the order of integration as \( dz \, dx \, dy \). Based on the boundary, the iterated integral becomes: \[\int_{y=0}^{4} \int_{x=0}^{\frac{1}{2}y} \int_{z=0}^{2} f(x, y, z) \, dz \, dx \, dy\]This order respects the bounds for each variable set by the region \( S \).

Sketch the Solid

Draw a triangular region on the \( xy \)-plane where \( 0 \leq x \leq \frac{1}{2}y \) and \( 0 \leq y \leq 4 \). Extend this triangular region vertically from \( z = 0 \) to \( z = 2 \) to visualize the volume of the solid \( S \).

Key concepts

Iterated Integral

An iterated integral is a way to calculate integrals over multidimensional regions by integrating one variable at a time. In our exercise, we are working with a triple integral. This involves three integrals stacked upon one another - one for each variable: in this case, x, y, and z.
Choosing the correct order of integration is crucial. It depends on the given boundaries and the nature of the relationships between the variables. In the problem, the boundaries suggest using the order: \( dz \, dx \, dy \). This sequence fits nicely with the constraints for each variable.
To evaluate an iterated integral, you start by integrating with respect to one variable while treating others as constants, moving outwards until all the integrals are evaluated. This method simplifies complex multi-variable integration problems.

Volume of Solid

Understanding the volume of a solid through integration is an essential concept in calculus. In our exercise, the goal is to use a triple integral to determine the volume of a solid shape defined by specific bounds.
The iterative process of integrating in the sequence \( dz \, dx \, dy \) accumulates the small differential volumes \( dV \), which together sum up to the total volume of the solid. Given the nature of the bounding parameters \( 0 \leq x \leq \frac{1}{2}y \), \( 0 \leq y \leq 4 \), and \( 0 \leq z \leq 2 \), our approach divides the solid's volume into tiny segments, integrating them over the region S.
This process succinctly helps in calculating the volume, turning a seemingly complex shape into manageable mathematical segments.

Boundaries in Coordinate Systems

Recognizing and interpreting boundaries in coordinate systems enables us to outline the shape of a region accurately in three-dimensional space. In this exercise, boundaries are given by \( 0 \leq x \leq \frac{1}{2}y \), \( 0 \leq y \leq 4 \), and \( 0 \leq z \leq 2 \).
Understanding these constraints is key in carving out the correct region to be integrated. The boundary \( x = \frac{1}{2}y \) defines a linear relationship between x and y, which forms the diagonal of a triangle in the \( xy \)-plane. This nested approach ensures each boundary condition fits seamlessly without overlap or gaps.
Visualizing these boundaries helps in correctly setting up the integral, providing the limits of integration for each variable and ensuring they collectively cover the entire region without over or underestimating.

3D Visualization in Calculus

3D visualization is invaluable in calculus, especially when dealing with triple integrals and solid volumes. Being able to mentally picture the solid shape bounded by the inequalities and described in the integration problem makes solving it much easier.
The exercise asks us to visualize the 3D region by extending a triangular shape in the \( xy \)-plane into a solid from \( z = 0 \) to \( z = 2 \). This prism-like shape emerges naturally from the constraints on z, bounded vertically to form a volume consistent with our bounding conditions.
With 3D visualization, we translate mathematical conditions into tangible shapes, making it possible to better understand and solve problems involving complex volumes.